Proof of Threesigned Equivalence to Complex Numbers
Products in threesigned math exactly match products in complex arithmetic.
Sums are also equivalent.
By choosing the star pole to be on the positive real axis the product arithmetic
works out. The transform from a threesigned value to a complex value is:
z = s(y)  (1/2)(m(y) + p(y)) + i(sqrt(3)/2)(m(y)  p(y))
Where y is in P3 and m(),p(),s() return the minus,plus and star
components of their argument.
These equations come from pure graphical analysis of the threesigned
branches being at angles 2pi/3. The components resolve via right
triangles. The minus pole is oriented toward the positive imaginary axis.
 minus pole

 +Imaginary
 
 
 
 .  . . . . .p1 = * 7  4
  .
  .
.  .
. 0 * * * * * * * * * * star pole, +Real
. + origin .
p3 = + 4  2 . + .
. + .
+ .
+ .
+ . . . . . . . . . p2 = * 9 + 6
+
+
+
+ plus pole
It will be shown that
z( y1 y2 ) = z( y1 ) z( y2 )
where the product on the left is performed in P3 and the product on the right is performed in C.
To simplify notation I will simply use s,m,and p to represent the
three sign components. I will also use q3 to represent the square root of
three.
We have
z1 = s1(1/2)(m1+p1) + i(q3/2)(m1p1)
and
z2 = s2(1/2)(m2p2) + i(q3/2)(m2p2) .
The complex product z1z2 is then:
(s1m1/2p1/2 + iq3m1/2iq3p1/2)(s2m2/2p2/2 + iq3m2/2iq3p2/2)
= s1s2 s1m2/2 s1p2/2 +s1m2iq3/2s1p2iq3/2
+ m1s2/2 +m1m2/4 +m1p2/4 m1m2iq3/4 +m1p2iq3/4
+ p1s2/2 +p1m2/4 +p1p2/4 p1m2iq3/4 +p1p2iq3/4
+ +m1s2iq3/2 m1m2iq3/4 m1p2iq3/4 m1m2(3/4) +m1p2(3/4)
+ p1s2iq3/2 +p1m2iq3/4 +p1p2iq3/4 +p1m2(3/4) p1p2(3/4) .
Gathering terms with precedence in the order of s,m,p:
= s1s2 s1m2/2 +s1m2iq3/2 s1p2/2 s1p2iq3/2
+ m1s2/2 +m1s2iq3/2 +m1m2/4 m1m2(3/4) m1m2iq3/4 m1m2iq3/4 +m1p2/4
+m1p2iq3/4 m1p2iq3/4 +m1p2(3/4)
+ p1s2/2 p1s2iq3/2 +p1m2/4 +p1m2(3/4) +p1m2iq3/4 p1m2iq3/4 +p1p2/4
p1p2(3/4) +p1p2iq3/4 +p1p2iq3/4
and reducing cancellable terms:
= s1s2 s1m2/2 +s1m2iq3/2 s1p2/2 s1p2iq3/2
+ m1s2/2 +m1s2iq3/2 m1m2/2 m1m2iq3/2 +m1p2
+ p1s2/2 p1s2iq3/2 +p1m2 p1p2/2 +p1p2iq3/2
[ Equation 1 ] .
The product y1y2 has the following properties based on its definition:
m = m1s2 + p1p2 + s1m2.
p = m1m2 + p1s2 + s1p2.
s = m1p2 + p1m2 + s1s2.
Converted to a complex value it is:
+m1p2 +p1m2 +s1s2 (1/2){m1s2 +p1p2 +s1m2 +m1m2 +p1s2 +s1p2 }
+i(q3/2){m1s2 +p1p2 +s1m2 (m1m2 + p1s2 + s1p2)}
= m1p2 +p1m2 +s1s2 m1s2/2 p1p2/2 s1m2/2 m1m2/2 p1s2/2 s1p2/2
+m1s2iq3/2 +p1p2iq3/2 +s1m2iq3/2 m1m2iq3/2 p1s2iq3/2  s1p2iq3/2 .
Now gathering terms from the above to better match the complex product
of Equation 1:
= +s1s2 s1m2/2 +s1m2iq3/2 s1p2/2  s1p2iq3/2
m1s2/2 +m1s2iq3/2 m1m2/2 m1m2iq3/2 +m1p2
p1s2/2 p1s2iq3/2 +p1m2 p1p2/2 +p1p2iq3/2 .
This exactly matches the reduced form of the complex product and
represents the equation of the product of two threesigned values y1y2
(where s1,m1,p1 represents y1 and s2,m2,p2 represents y2) converted to
the traditional complex representation.
Sums are also equivalent.
The proof is trivial and is already apparent from the graphical behavior.
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